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Consider the left shift operator $L:\mathbb{R}^\infty\to\mathbb{R}^\infty$ given by

$$ L(x_1,x_2,x_3,\ldots)=(x_2,x_3,\ldots). $$

Does the dual operator $L^\prime:(\mathbb{R}^\infty)^\prime\to(\mathbb{R}^\infty)^\prime$ have an eigenvector?

I want to say no via the following argument. An eigenvector $\varphi\in\left(\mathbb{R}^{\infty}\right)^{\prime}$ of $L^{\prime}$ would need to satisfy the eigenvalue equation $L^{\prime}\varphi=\alpha\varphi=\varphi\circ L$ for some $\alpha\in\mathbb{R}$. Applying this functional identity to an arbitrary element of $\mathbb{R}^{\infty}$ gives $$\alpha\varphi\left(x_{1},x_{2},x_{3},\ldots\right)=\varphi\left(x_{2},x_{3},x_{4},\ldots\right).$$

If we vary $x_{1}$, we see that the argument of the LHS of the above equation varies, but the argument of the RHS does not. Thus for our putative eigenvector $\varphi$ to satisfy the above functional identity, we require that $\varphi$ does not depend on the variable in the first slot of its argument. But this means that if we vary $x_{2}$, then the argument of the RHS of the above equation must remain unchanged, since $x_{2}$ is in the first slot of the argument of $\varphi$ in the RHS. But varying $x_{2}$ changes the second slot of the argument of the LHS, so we conclude that the above equation is only consistent if $\varphi$ does not depend on the variable in the second slot of its argument. Repeating this argument ad infinitum tells us that $\varphi$ must be independent of its argument entirely and hence a constant function $\varphi=c$. But the only constant functional that is also linear is $\varphi=0$, which cannot be an eigenvector since eigenvectors are defined to be nonzero. Hence $L^{\prime}$ has no eigenvectors.

After some thought however, I realized a flaw with this argument - namely, there could potentially exist a linear function which remains constant if I change a finite number of its arguments (what I proved) but nonetheless changes value if I simultaneously change an infinite number of its arguments. One such example that is non-linear is

$$ \varphi\left(x_{1},x_{2},\ldots\right)=\begin{cases} \lim_{n\to\infty}x_{n}, & \text{if this limit exists,}\\ 0, & \text{otherwise.} \end{cases} $$

My argument wouldn't rule out a function like this, but any such function I try to come up with ends up being non-linear.

Is it possible to tweak my argument by applying linearity so that I can rule out such pathological possibilities? Or does $L^{\prime}$ actually possess such a pathological, but linear, eigenvector (constructed perhaps via the axiom of choice)?

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  • $\begingroup$ To be clear, by $\mathbb{R}^\infty$ you mean the vector space of all sequences, and $(\mathbb{R}^\infty)'$ is its algebraic dual? $\endgroup$ – Eric Wofsey Oct 1 '17 at 1:59
  • $\begingroup$ Yes, precisely. $\endgroup$ – sferics Oct 1 '17 at 2:45
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Note that for any sequence $x=(x_1,x_2,x_3,\dots)$, there is another sequence $y=(y_1,y_2,y_3,\dots,)$ such that $x=\alpha y-Ly$. Namely, let $y_1$ be anything, and then recursively define $y_{n+1}=\alpha y_n-x_n$. If $\varphi$ is an eigenvector of $L'$ with eigenvalue $\alpha$ existed, then $\alpha\varphi(y)=\varphi(Ly)$ and so $0=\varphi(\alpha y-Ly)=\varphi(x)$. Since $x$ was arbitrary, this means $\varphi=0$.

To put it another way, the map $\alpha I-L$ is surjective, which implies its dual map $\alpha I-L'$ is injective, so $\alpha$ cannot be an eigenvalue of $L'$.

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  • $\begingroup$ Simple, I like it! Thanks! $\endgroup$ – sferics Oct 1 '17 at 8:54

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